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# PPOOOPP geometry

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In the figure, if $MR=MK$, the measure of arc $MK$ is 130 degrees, and measure of arc $MQ$ is 28 degrees, then what is $\angle RPK$, in degrees?

#1
+4476
+3

Since   $$MR = MK$$   and   $$\overset{\frown}{MK} = 130°$$   ,   $$\overset{\frown}{MR} = 130°$$

$$\overset{\frown}{MK} +\overset{\frown}{KR} +\overset{\frown}{MR} =360°$$

Replace  $$\overset{\frown}{MK}$$  with  130°  and  $$\overset{\frown}{MR}$$  with  130°

$$130°+\overset{\frown}{KR}+130°=360° \\~\\ \overset{\frown}{KR}=360°-130°-130° \\~\\ \overset{\frown}{KR}=100°$$

And..

$$m\angle RPK=\frac{\overset{\frown}{KR}+\overset{\frown}{MQ}}{2}\\~\\ m\angle RPK=\frac{100°+28°}{2} \\~\\ m\angle RPK=64°$$

hectictar  Aug 26, 2017
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#1
+4476
+3

Since   $$MR = MK$$   and   $$\overset{\frown}{MK} = 130°$$   ,   $$\overset{\frown}{MR} = 130°$$

$$\overset{\frown}{MK} +\overset{\frown}{KR} +\overset{\frown}{MR} =360°$$

Replace  $$\overset{\frown}{MK}$$  with  130°  and  $$\overset{\frown}{MR}$$  with  130°

$$130°+\overset{\frown}{KR}+130°=360° \\~\\ \overset{\frown}{KR}=360°-130°-130° \\~\\ \overset{\frown}{KR}=100°$$

And..

$$m\angle RPK=\frac{\overset{\frown}{KR}+\overset{\frown}{MQ}}{2}\\~\\ m\angle RPK=\frac{100°+28°}{2} \\~\\ m\angle RPK=64°$$

hectictar  Aug 26, 2017
#2
+76094
+2

Draw RK

Since minor arc MK = 130°....then angle KRM  = (1/2)  of this = 65°

But  MR  = MK...so angle RKM  also  = 65°

And since minor arc MQ = 28°...then angle QKM  = (1/2)  of this = 14°

Therefore, angle RKP  =  m< RKM - m< QKM  = 65 - 14  = 51°

Then, in triangle RKP....angle KRP = angle KRM  = 65°

And angle RKP  = 51°

Therefore, angle  RPK  = 180 - 65 - 51  = 64°

CPhill  Aug 26, 2017
edited by CPhill  Aug 26, 2017
#3
+76094
+1

Rats...!!!...hectictar beat me to it!!!

Her method is better, anyway.....no lines had to be drawn and she just dealt with arc measures....much more efficient.....

CPhill  Aug 26, 2017
#4
+4476
+3

Hahah!! Well, maybe it's a little more efficient, but it isn't as explanatory  :)

hectictar  Aug 26, 2017
edited by hectictar  Aug 26, 2017
#7
+675
+1

Lancelot described many of Heureka’s works as “Songs Without Words.”

This also describes some of your presentations. The others are “Songs With Words.”

GingerAle  Aug 26, 2017
#5
+76094
+2

Here's one other method to solve this  :

Since chord KM = chord RM....then  arc KMR =  130 + 130  = 260°

So....minor arc RK  = 360 - 260  = 100°

And angle RMK is an inscribed angle subtending this arc...so its measure  = 50°

And angle QKM subtends minor arc QM = 28°....so  angle QKM  = 14°

And by  the exterior angle theorem angle RPK  = RMK + QKM   = 50 + 14  = 64°

[ Note...in essence....this is the same proof as hectictar's   ....!!!  ]

CPhill  Aug 26, 2017
edited by CPhill  Aug 26, 2017
#6
+4476
+3

Hey...here's another way!

∠KRM  =  130° / 2  =  65°

∠RKM  =  130° / 2  =  65°

∠QKM  =  28° / 2  =  14°

∠RKQ  =  65° - 14°  =  51°

And since there are 180° in triangle KPR....     ∠RPK  =  180° - 65° - 51°  =  64°

*edit*

Wait a minute...this isn't any different than the second one!!! Ah..sorry about that!

hectictar  Aug 26, 2017
edited by hectictar  Aug 26, 2017
#8
+675
+1

I would describe this as your variation on a theme by Sir CPhill.

GingerAle  Aug 26, 2017

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